题目内容
1.已知tanα=3,计算:(1)5cosα+3sinα;
(2)sinαcosα.
分析 利用弦化切思想,结合已知,
(1)先求(5cosα+3sinα)2=25cos2α+30sinαcosα+9sin2α=$\frac{25{cos}^{2}α{+30sinα•cosα+9sin}^{2}α}{{sin}^{2}α+{cos}^{2}α}$=$\frac{25{+30tanα+9tan}^{2}α}{{tan}^{2}α+1}$的值,可求(1)的值;
(2)将sinαcosα=$\frac{{sinα•cos}^{\;}α}{{sin}^{2}α+{cos}^{2}α}$=$\frac{{tanα}^{\;}}{{tan}^{2}α+1}$可求(2)的值;
解答 解:∵tanα=3,
(1)(5cosα+3sinα)2=25cos2α+30sinαcosα+9sin2α=$\frac{25{cos}^{2}α{+30sinα•cosα+9sin}^{2}α}{{sin}^{2}α+{cos}^{2}α}$=$\frac{25{+30tanα+9tan}^{2}α}{{tan}^{2}α+1}$=$\frac{25+90+81}{10}$=$\frac{196}{10}$=$\frac{98}{5}$,
∴5cosα+3sinα=$\frac{7\sqrt{10}}{5}$,或5cosα+3sinα=-$\frac{7\sqrt{10}}{5}$,
(2)sinαcosα=$\frac{{sinα•cos}^{\;}α}{{sin}^{2}α+{cos}^{2}α}$=$\frac{{tanα}^{\;}}{{tan}^{2}α+1}$=$\frac{3}{10}$.
点评 本题考查的知识点是同角三角函数的基本关系的运算--弦化切思想,难度中档.
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围是( )
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